Problem: The line $y = \frac{5}{3} x - \frac{17}{3}$ is to be parameterized using vectors.  Which of the following options are valid parameterizations?

(A) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} -3 \\ -5 \end{pmatrix}$

(B) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 17 \\ 5 \end{pmatrix} + t \begin{pmatrix} 6 \\ 10 \end{pmatrix}$

(C) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -7/3 \end{pmatrix} + t \begin{pmatrix} 3/5 \\ 1 \end{pmatrix}$

(D) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 14/5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 3/5 \end{pmatrix}$

(E) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ -17/3 \end{pmatrix} + t \begin{pmatrix} 15 \\ -25 \end{pmatrix}$

Enter the letters of the correct options, separated by commas.
Explanation: Note that $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ are two points on this line, so a possible direction vector is
\[\begin{pmatrix} 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}.\]Then any nonzero scalar multiple of $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ can also be a direction vector.

The form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v} + t \mathbf{d}\]parameterizes a line if and only if $\mathbf{v}$ lies on the line, and $\mathbf{d}$ is a possible direction vector for the line.  Checking, we find that the possible parameterizations are $\boxed{\text{A,C}}.$